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At equilibrium, the concentrations of N and D are equal, meaning $[N] = [D]$. The equilibrium constant $K_{eq}$ is defined as: $$K_{eq} = \frac{[D]}{[N]} = 1$$
The standard Gibbs free energy change $\Delta G^{\circ}$ is related to $K_{eq}$ by the equation: $$\Delta G^{\circ} = -RT \ln(K_{eq})$$ Since $K_{eq} = 1$, $\ln(1) = 0$, therefore $\Delta G^{\circ} = 0$.
We know that $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$. Given $\Delta G^{\circ} = 0$, we have: $$\Delta H^{\circ} = T\Delta S^{\circ}$$
Convert the temperature to Kelvin: $T = 60 + 273.15 = 333.15 K$. Given $\Delta H^{\circ} = 666 kJ mol^{-1}$, we solve for $\Delta S^{\circ}$: $$\Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} = \frac{666}{333.15} \approx 2 kJ K^{-1} mol^{-1}$$
Final Answer: 2